Explanation Given information: The figure shown below shows the free body diagram of the system Figure-(1) Write the expression for the rate change of momentum. H ˙ G = H G x y z + Ω × H G .........(I) Here, the rate of change of angular momentum is H ˙ G , angular momentum with respect to the frame of reference and angular velocity of the xyz frame is Ω . Write the expression for the angular momentum about the centre of mass. H G = I x ω i + I x y ω j + I x z ω k ……. (II) Here, momentum of inertia about x direction is I x , momentum of inertia about x y plane is I x y , momentum of inertia about x z plane is I x z and the angular velocity is x z . Substitute I x ω i + I x y ω j + I x z ω k for H G , ω i for Ω and 0 for H G x y z in Equation (I). H ˙ G = 0 + ω i × ( I x ω i + I x y ω j + I x z ω k ) = I x z ω j + I x y ω k ……. (III) The shaft lies between x z plane hence the value of I x y is 0 . Substitute 0 for I x y in Equation (III). H ˙ G = I x z ω 2 j + ( 0 ) ω 2 k = I x z ω 2 j .........(IV) The magnitude of A and B is Equal and opposite in direction because the center of mass lies on the line A B as it is symmetrical A and B forms a couple. A = − ( B ) .........(V) Write the expression for the force at B . F B = 4 a i × ( B y j + B z k ) = 4 a B y k − 4 a B z j .........(VI) The magnitude of B in the y direction is 0 . Substitute 0 for B y in Equation (VI). F B = 0 − 4 a B z j = − 4 a B z j .........(VII) Here, the magnitude of couple at B is F B the distance between a A to B is 4 a .the magnitude of B in z direction is B z

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Chapter 18.2, Problem 18.68P

To determine

The dynamic reaction at A.

The dynamic reaction at point B.

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Chapter 18.2, Problem 18.67P

Chapter 18.2, Problem 18.69P

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Given answers to be: i) 14.65 kN; 6.16 kN; 8.46 kN ii) 8.63 kN; 9.88 kN iii) Bearing 6315 for B1 & B2, or Bearing 6215 for B1

(b) A steel 'hot rolled structural hollow section' column of length 5.75 m, has the cross-section shown in Figure Q.5(b) and supports a load of 750 kN. During service, it is subjected to axial compression loading where one end of the column is effectively restrained in position and direction (fixed) and the other is effectively held in position but not in direction (pinned). i) Given that the steel has a design strength of 275 MN/m², determine the load factor for the structural member based upon the BS5950 design approach using Datasheet Q.5(b). [11] ii) Determine the axial load that can be supported by the column using the Rankine-Gordon formula, given that the yield strength of the material is 280 MN/m² and the constant *a* is 1/30000. [6] 300 600 2-300 mm wide x 5 mm thick plates. Figure Q.5(b) L=5.75m Pinned Fixed

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Chapter 18 Solutions

Vector Mechanics For Engineers

Ch. 18.1 - Prob. 18.1P Ch. 18.1 - Prob. 18.2P Ch. 18.1 - Prob. 18.3P Ch. 18.1 - A homogeneous disk of weight W=6 lb rotates at the...Ch. 18.1 - Prob. 18.5P Ch. 18.1 - A solid rectangular parallelepiped of mass m has a...Ch. 18.1 - Solve Prob. 18.6, assuming that the solid...Ch. 18.1 - Prob. 18.8P Ch. 18.1 - Determine the angular momentum HD of the disk of...Ch. 18.1 - Prob. 18.10P

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